Let $$\mathfrak{u}(n)=\{X\in M(n,\Bbb C):X+X^*=0\}$$ where $X^*$ is the conjugate transpose. Then, $\mathfrak{u}(n)$ is a real vector space.
Problem. Show that $\langle X,Y\rangle=\DeclareMathOperator{\Tr}{Tr}\Tr(XY^*)$ is a real inner-product on $\mathfrak{u}(n)$ that is invariant under conjugation by $U(n)$.
Attempts.
The part about conjugation by $U(n)$ is easy: $$\Tr((UXU^{-1})(UYU^{-1})^*)=\Tr(UXU^{-1}UY^*U^*)=\Tr(UXY^*U^{-1})=\Tr(XY^*)$$ since the trace is invariant under conjugation.
But now, I don't know how to prove that $\Tr(XY^*)$ is an inner-product.
It is clear every matrix in $\mathfrak{u}(n)$ is normal and hence unitarily diagonalizable. So $X\sim \DeclareMathOperator{\diag}{diag}\diag(a_1,\ldots,a_n)$ and $Y\sim \diag(b_1,\ldots,b_n)$ where $a_i,b_i\in\Bbb R$. My guess is that $$\Tr(XY^*)=a_1b_1+\cdots+a_nb_n,$$ and hence it is a real inner-product. This is true if $X$ and $Y$ are diagonal. But I can't prove it in general.